Problem: $f(n) = -n-5$ $h(n) = 2n-3(f(n))$ $g(x) = -6x-5(h(x))$ $ f(h(-1)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(-1)$ . Then we'll know what to plug into the outer function. $h(-1) = (2)(-1)-3(f(-1))$ To solve for the value of $h$ , we need to solve for the value of $f(-1)$ $f(-1) = -(-1)-5$ $f(-1) = -4$ That means $h(-1) = (2)(-1)+(-3)(-4)$ $h(-1) = 10$ Now we know that $h(-1) = 10$ . Let's solve for $f(h(-1))$ , which is $f(10)$ $f(10) = -10-5$ $f(10) = -15$